Pronormality is not finite-upper join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., pronormal subgroup) not satisfying a subgroup metaproperty (i.e., finite-upper join-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Upper join-closed subgroup property (?), .
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about pronormal subgroup|Get more facts about finite-upper join-closed subgroup propertyGet more facts about upper join-closed subgroup property|

Statement

We can have a subgroup $H\le G$ such that $H$ is a pronormal subgroup in two intermediate subgroups $K$ and $L$, but $H$ is not pronormal in the join $⟨K,L⟩$.

Related facts

Related facts about pronormality

• Pronormality satisfies intermediate subgroup condition: If $H\le K\le G$ and $H$ is pronormal in $G$, then $H$ is also pronormal in $K$.

Related facts about upper join-closedness

• 2-subnormality is not upper join-closed
• Subnormality is not upper join-closed
• Normality is upper join-closed
• Characteristicity is not upper join-closed

Proof

Example

Further information: symmetric group:S3, symmetric group:S4

Let $G$ be the symmetric group on the set $\{1,2,3,4\}$, $K$ be the subgroup comprising permutations on $\{1,2,3\}$, $L$ be the subgroup comprising permutations on $\{1,2,4\}$, and $H=K\cap L$ be the two-element subgroup generated by $(1,2)$. Observe that:

• $H$ is a pronormal subgroup in both $K$ and $L$. In fact, it is a maximal subgroup in both, and maximal implies pronormal.
• $⟨K,L⟩=G$.
• $H$ is not pronormal in $G$. That's because the subgroup generated by $(3,4)$ is a conjugate of $H$, and these two subgroups are not conjugate in the subgroup they generate.